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STATICS / STRENGTH OF MATERIALS -Example

In the structure shown on right horizontal member BCD is supported by vertical brass member, AB, an aluminum member DE, and by a roller at point C. Both AB and ED have cross sectional areas of .5 in².
The structure is initially unstressed and then experiences a temperature increase of 60 degrees Celsius. For this structure:
A. Draw a Free Body Diagram showing all support forces and loads.
B. Determine the axial stress that develops in aluminum member DE.
C. Determine the resulting movement of point B .
E_st = 30 x 10 ⁶ psi; E_br = 15 x 10 ⁶ psi; E_al = 10 x 10 ⁶ psi
a_st = 12 x 10^-6 /^oC; a_br = 20 x 10^-6 /^oC; a_al = 23 x 10^-6 /^oC
Unless otherwise indicated, all joints and support points are assumed to be pinned or hinged joints.

Solution:

PART A: STATICS

STEP 1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions.

We first examine how much the members would expand freely, due to temperature increase, if they were free to expand (Diagram 2).

Since both members can not expand as shown (assuming member BCD does not bend), one member will "win," expanding and compression the other member. We will assume the brass member actually expands, compression the aluminum member (Diagram 3).

Since the two members are pushing against each other, both members are put into compression, and the support reaction are as shown in diagram 3.
Now we write the equilibrium equations:
Sum F_x = 0 (None)
Sum F_y = -F_BR + F_C - F_AL = 0
Sum T_C = F_BR (4 ft) - F_AL (12 ft) = 0
Not enough independent equations to solve for our three unknowns, so we need another equation - which we will obtain from the deformation relationship.

PART B: DEFORMAION
General relationship from geometry of structure:
d_BR / 4 ft = - d_AL / 12 ft ; or 3 d_BR = - d_AL
(negative sign is due to our assumption that the aluminum got shorter (is compressed).)

We now expand the expression for our deformations using the fact that the total deformation is given by:
d_total = [a DTL ⁺ FL/EA]_material so,
3 [a DTL - FL/EA]_brass = - [a DTL - FL/EA]_aluminum

or, using known values:
3[ (20x10^-6/^oC) (+60^oC) (96 in) - F_BR (96 in) / (15x10⁶ lbs/in²) (.5 in²) ] =
[ (23x10^-6/^oC) (+60^oC) (48 in) - F_AL (48 in) / (10x10⁶ lbs/in²) (.5 in²) ]
or + (0.346 - 3.84x10^-5 F_BR) = - (0.0662 - 0.96x10^-5 F_AL) or
3.84x10^-5 F_BR + 0.96x10^-5 F_AL = 0.412 in

From our statics, we had
F_BR (4 ft) - F_AL (12 ft) = 0 or F_BR = 3 F_AL,
which we now substitute into our deformation relationship
3.84x10^-5 (3 F_AL) + 0.96x10^-5 F_AL = 0.412 in
Solving for
F_AL = 3,330 lbs ; F_BR = 9,900 lbs ; s_AL = F/A = 3,300 lbs/.5 in² = 6,600 lbs/in²

PART C: MOVEMENT
Mov. B = the deformation of member AB
So, Mov. B = [(20x10^-6 / ^oC)(60^oC)(96 in) - (9,900 lbs)(96 in) / (15x10⁶ lbs/in² )(.5 in² )]
Mov. B = -.01152" (This indicates that the brass member is compressed this amount.)